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I know when we talk about limit of function ,there should be with respect to topology on both domain and codomain. And sequence from topological space are basically function from set of natural number to that topological space.My question is. (1) which topology we assume on set of natural when we are Finding limit of sequence?

MJD
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MEET PATEL
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    We usually don’t assume topologies on the natural numbers (though this can be done). The limit of a sequence is usually just defined by a topology in the codomain; the only burden on the domain is that “for all large $n$, the sequence $a_n$ stays in the neighbourhood” – FShrike Jul 26 '22 at 14:58
  • @FShrike cant we use order topology (usual order relation)on set of extended natural number? – MEET PATEL Jul 26 '22 at 15:00
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    We can, but we don't necessarily have to. What FShrike is saying (or at least my understanding of what they are staying) is that there isn't a "default" topology that we assume on the natural numbers, so any problem involving continuity should specify the underlying topology. – Connor Gordon Jul 26 '22 at 15:07

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By default we don't place any topology on $\mathbb{N}$ when we talk about limits of sequences. If you really want to talk about limits of sequences in this language you can do the following. Take the extended natural numbers $\mathbb{N}_{\infty} = \mathbb{N} \cup \{ \infty \}$ together with the topology generated by the discrete topology on $\mathbb{N}$ and the open sets $\{ n \ge N \} \cup \{ \infty \}$ for all $N$ (this is the one-point compactification of $\mathbb{N}$). Equivalently, this is the subspace topology given by the embedding

$$\mathbb{N}_{\infty} \ni n \mapsto \frac{1}{n} \in [0, 1]$$

where $\frac{1}{\infty} = 0$.

With respect to this topology, finding the limit as $n$ goes to $\infty$ of a sequence of elements of a Hausdorff topological space $X$ is equivalent to finding a continuous extension of this sequence from a function $\mathbb{N} \to X$ to a function $\mathbb{N}_{\infty} \to X$. If such a continuous extension exists, the two definitions of a limit (the usual one for sequences vs. the one for functions between topological spaces) agree, and both agree with evaluation at $\infty$.

Qiaochu Yuan
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  • For Embedding u show above Can we say it is one point compactification of set of natural number? , – MEET PATEL Jul 26 '22 at 15:13
  • And in continuous extension we should assign \infty to limit of sequence in \mathbb{N}{\infty} \to X Right? Well if it is then topology on {N}{\infty} when we r talking about continuous extension Is as you above mentioned? – MEET PATEL Jul 26 '22 at 15:19
  • Yes, it's the one-point compactification. And yes, that's what I said. – Qiaochu Yuan Jul 26 '22 at 15:20
  • If space is not Hausdorff then sequence may not have unique limit ,then what happen to the extension ? I mean if sequence has for example 2 limits then in either case that extension will continuous? – MEET PATEL Jul 26 '22 at 15:28
  • Yes, continous extensions correspond bijectively to limits, so if limits aren't unique then continuous extensions aren't unique either (but are still continuous). – Qiaochu Yuan Jul 26 '22 at 15:32
  • And You said "the two definitions of a limit (the usual one for sequences vs. the one for functions between topological spaces) agree, " plz bit more on this.. – MEET PATEL Jul 26 '22 at 15:33
  • Can you ask a more specific question? This is a good exercise and you should try to do it yourself if you can. – Qiaochu Yuan Jul 26 '22 at 15:35
  • Asking it bcz ambiguity of two defination.i only know is sequence in topological space converges to point a if,for every open set V of a, terms of sequences are eventually in V.then what u mean by second definition? – MEET PATEL Jul 26 '22 at 15:39
  • Isn't this exactly what you were talking about when you said "I know when we talk about limit of function ,there should be with respect to topology on both domian and codomain."? I mean the definition in terms of open sets: https://math.stackexchange.com/a/3151525/232 – Qiaochu Yuan Jul 26 '22 at 15:42